Recent content by derek101

I don't understand,why if 10% of electrons were to pass through a barrier,must I assume that the electron has a 10% chance of tunnelling it's way through. Why can't I assume that the barrier only has a 90% chance of stopping the electrons?

If an electron and positron are considered as 2 opposites about a point of equilibrium,they can head back to the point of equilibrium,but energy must be conserved,so photons can be considered as oscillations about the point of equilibrium,a photon being it's own anti-particle. Electron/positron...

Think of it as 2 opposites,about a point of equilibrium.

Where would the energy come from?

Question.So does the model I described work ok for gas and liquids? A photon entering a solid is no longer a particle until it is re-emitted/re-created.

When photons travel through a medium they travel at the speed of light,but their rate of propagation is slowed by being absorbed and re-emitted by atoms.

relativistic aberration would change your view of the universe.

Should I be thinking that the macroscopic world and It's laws of physics,emerge from a totally different set of physical laws of the microscopic quantum world. Or is this the wrong approach?

If the the repulsive force between 2 protons,is mediated by the creation and exchange of virtual particles,and this field of virtual particles is the barrier,then you have just as much uncertainty about the position and strength of the barrier,as you do about the position and momentum of the...

Drakkith If it's the particle itself that has the uncertainty then:- Are you saying that 2 protons in the vacuum of space,have the same chance of tunneling through the barrier,as 2 protons in the middle of the sun?

It's not that the particle has enough energy to get through the barrier,energy fluctuations in the barrier can push the particle through it(by absorbing and emitting).

ok thanks for link. not quiet sure how I use the algorithm,but see I can make mirror image. bitmap(200+x,200-y)=1 bitmap(200+x,200+y)=1 bitmap(200-x,200-y)=1 bitmap(200-x,200+y)=1 plus mirror image bitmap(200+y,200-x)=1 bitmap(200+y,200+x)=1 bitmap(200-y,200-x)=1 bitmap(200-y,200+x)=1for x = 0...

e.g:- rounding of 10.457 to the nearest 2 decimal places (integer(10.457*100+0.5))/100 = 1046/100 = 10.46 for 3 decimal places *1000 and then /1000

ah sussed it I can use Pythagoras. dim bitmap(400,400) as integer radius=100 z=radius*radius for x=0 to radius t=x*x d=z-t y=d^(1/2) bitmap(200+x,200-y)=1 bitmap(200+x,200+y)=1 bitmap(200-x,200-y)=1 bitmap(200-x,200+y)=1 next xonly problem is this does not give a continuous line of points on...